Solution

Since the game is a home game, let's again suppose that 80% of the fans attending the game are Penn State fans, while 20% are Notre Dame fans. That is, \(P(P) = 0.8\) and \(P(N) = 0.2\). Then, by independence:

\(P(X = 0) = P(NNN) = 0.2 \times 0.2 \times 0.2 = 1 \times (0.8)^0\times (0.2)^3\)

And, by independence and mutual exclusivity of \(NNP\), \(NPN\), and \(PNN\):

\(P(X = 1) = P(NNP) + P(NPN) + P(PNN) = 3 \times 0.8\times 0.2\times 0.2 = 3\times (0.8)^1\times (0.2)^2\)

Likewise, by independence and mutual exclusivity of \(PPN\), \(PNP\), and \(NPP\):

\(P(X = 2) = P(PPN) + P(PNP) + P(NPP) = 3\times 0.8 \times 0.8 \times 0.2 = 3\times (0.8)^2\times (0.2)^1\)

Finally, by independence:

\(P(X = 3) = P(PPP) = 0.8\times 0.8\times 0.8 = 1\times (0.8)^3\times (0.2)^0\)

Do you see a pattern in our calculations? It seems that, in each case, we multiply the number of ways of obtaining \(x\) Penn State fans first by the probability of \(x\) Penn State fans \((0.8)^x\) and then by the probability of \(3-x\) Nebraska fans \((0.2)^{3-x}\).

Answer

Yes, \(X\) is a binomial random variable, because:

1. The coin is tossed in exactly the same way 100 times.
2. Each toss results in either a head (success) or a tail (failure).
3. One toss doesn't affect the outcome of another toss. The trials are independent.
4. The probability of getting a head is 0.70 for each toss of the coin.
5. \(X\) equals the number of heads (successes).

Answer

No, \(X\) is not a binomial random variable, because the number of trials \(n\)was not fixed in advance, and \(X\) does not equal the number of volunteers in the sample.

Answer

No, \(X\) is not a binomial random variable, because \(p\), the probability that a randomly selected skein has acceptable color changes from trial to trial. For example, suppose, unknown to the QCI, that 9 of the 15 skeins of yarn in the lot are acceptable. For the first trial, \(p\)equals \(\frac{9}{15}\). However, for the second trial, \(p\)equals either \(\frac{9}{14}\) or \(\frac{8}{14}\)depending on whether an acceptable or unacceptable skein was selected in the first trial. Rather than being a binomial random variable, \(X\) is a hypergeometric random variable. If we continue to assume that 9 of the 15 skeins of yarn in the lot are acceptable, then \(X\) has the following probability mass function:

\(f(x)=P(X=x)=\dfrac{\dbinom{9}{x} \dbinom{6}{5-x}}{\dbinom{15}{5}}\) for \(x=0, 1, \ldots, 5\)

Answer

No, \(X\) is technically a hypergeometric random variable, not a binomial random variable, because, just as in the previous example, sampling takes place without replacement. Therefore, \(p\), the probability of selecting an SUV owner, has the potential to change from trial to trial. To make this point concrete, suppose that Americans own a total of \(N=270,000,000\) cars. Suppose too that half (135,000,000) of the cars are SUVs, while the other half (135,000,000) are not. Then, on the first trial, \(p\)equals \(\frac{1}{2}\) (from 135,000,000 divided by 270,000,000). Suppose an SUV owner was selected on the first trial. Then, on the second trial, \(p\) equals 134,999,999 divided by 269,999,999, which equals.... punching into a calculator... 0.499999... Hmmmmm! Isn't that 0.499999... close enough to \(\frac{1}{2}\) to just call it \(\frac{1}{2}\)?Yes...that's what we do!

In general, when the sample size \(n\)is small in relation to the population size \(N\), we assume a random variable \(X\), whose value is determined by sampling without replacement, follows (approximately) a binomial distribution. On the other hand, if the sample size \(n\)is close to the population size \(N\), then we assume the random variable \(X\) follows a hypergeometric distribution.

Solution

Since \(n=15\) is small relative to the population of \(N\) = 300,000,000 Americans, and all of the other criteria pass muster (two possible outcomes, independent trials, ....), the random variable \(X\) can be assumed to follow a binomial distribution with \(n=15\) and \(p=0.2\). Using the probability mass function for a binomial random variable, the calculation is then relatively straightforward:

\(P(X=3)=\dbinom{15}{3}(0.20)^3 (0.80)^{12}=0.25\)

That is, there is a 25% chance, in sampling 15 random Americans, that we would find exactly 3 that had no health insurance.

Solution

"At most one" means either 0 or 1 of those sampled have no health insurance. That is, we need to find:

\(P(X\le 1)=P(X=0)+P(X=1)\)

Using the probability mass function for a binomial random variable with \(n=15\) and \(p=0.2\), we have:

\(P(X \leq 1)=\dbinom{15}{0}(0.2)^0 (0.8)^{15}+ \dbinom{15}{1}(0.2)^1(0.8)^{14}=0.0352+0.1319=0.167\)

That is, we have a 16.7% chance, in sampling 15 random Americans, that we would find at most one that had no health insurance.

Solution

Yikes! "More than seven" in the sample means 8, 9, 10, 11, 12, 13, 14, 15. As the following picture illustrates, there are two ways that we can calculate \(P(X>7)\):

We could calculate \(P(X>7)\) by adding up \(P(X=8), P(X=9)\), up to \(P(X=15)\). Alternatively, we could calculate \(P(X>7)\) by finding \(P(X\le 7)\) and subtracting it from 1. But to find \(P(X\le 7)\), we'd still have to add up \(P(X=0), P(X=1)\) up to \(P(X=7)\). Either way, it becomes readily apparent that answering this question is going to involve more work than the previous two questions. It would clearly be helpful if we had an alternative to using the binomial p.m.f. to calculate binomial probabilities. The alternative typically used involves cumulative binomial probabilities.

Solution

The probability that at most 1 has no health insurance can be written as \(P(X\le 1)\). To find \(P(X\le 1)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 1 in the second column on the left, since we want to find \(F(1)=P(X\le 1)\).

Now, all we need to do is read the probability value where the \(p=0.20\) column and the (\(n=15, x=1\)) row intersect. What do you get?

We've used the cumulative binomial probability table to determine that the probability that at most 1 of the 15 sampled has no health insurance is 0.1671. For kicks, since it wouldn't take a lot of work in this case, you might want to verify that you'd get the same answer using the binomial p.m.f.

Solution

As we determined previously, we can calculate \(P(X>7)\) by finding \(P(X\le 7)\) and subtracting it from 1:

The good news is that the cumulative binomial probability table makes it easy to determine \(P(X\le 7)\) To find \(P(X\le 7)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 7 in the second column on the left, since we want to find \(F(7)=P(X\le 7)\).

Now, all we need to do is read the probability value where the \(p=0.20\) column and the (\(n = 15, x = 7\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(X\le 7)=0.9958\). Therefore:

\(P(X>7) = 1 − 0.9958 = 0.0042\)

That is, the probability that more than 7 in a random sample of 15 would have no health insurance is 0.0042.

Solution

We can calculate \(P(X=3)\) by finding \(P(X\le 2)\) and subtracting it from \(P(X\le 3)\), as illustrated here:

To find \(P(X\le 2)\) and \(P(X\le 3)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 3 in the second column on the left, since we want to find \(F(3)=P(X\le 3)\). And, find the 2 in the second column on the left, since we want to find \(F(2)=P(X\le 2)\).

Now, all we need to do is (1) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 3\)) row intersect, and (2) read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 2\)) row intersect. What do you get?

The cumulative binomial probability table tells us that finding \(P(X\le 3)=0.6482\) and \(P(X\le 2)=0.3980\). Therefore:

\(P(X = 3) = P(X \le 3) - P(X\le 2) = 0.6482-0.3980 = 0.2502\)

That is, there is about a 25% chance that exactly 3 people in a random sample of 15 would have no health insurance. Again, for kicks, since it wouldn't take a lot of work in this case, you might want to verify that you'd get the same answer using the binomial p.m.f.

Solution

We can calculate \(P(X\ge 1)\) by finding \(P(X\le 0)\) and subtracting it from 1, as illustrated here:

To find \(P(X\le 0)\) using the binomial table, we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 0 in the second column on the left, since we want to find \(F(0)=P(X\le 0)\).

Now, all we need to do is read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 0\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(X\le 0)=0.0352\). Therefore:

\(P(X\le 1) = 1-0.0352 = 0.9648\)

That is, the probability that at least one person in a random sample of 15 would have no health insurance is 0.9648.

Solution

"Fewer than 5" means 0, 1, 2, 3, or 4. That is, \(P(X<5)=P(X\le 4)\), and \(P(X\le 4)\) can be readily found using the cumulative binomial table. To find \(P(X\le 4)\), we:

1. Find \(n=15\) in the first column on the left.
2. Find the column containing \(p=0.20\).
3. Find the 4 in the second column on the left, since we want to find \(F(4)=P(X\le 4)\).

Now, all we need to do is read the probability value where the \(p = 0.20\) column and the (\(n = 15, x = 4\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(X\le 4)= 0.8358\). That is, the probability that fewer than 5 people in a random sample of 15 would have no health insurance is 0.8358.

Solution

If we let \(X\) denote the number of subscribers who qualify for favorable rates, then X is a binomial random variable with \(n=10\) and \(p=0.70\). And, if we let \(Y\) denote the number of subscribers who don't qualify for favorable rates, then \(Y\), which equals \(10-X\), is a binomial random variable with \(n=10\) and \(q=1-p=0.30\). We are interested in finding \(P(X\ge 4)\). We can't use the cumulative binomial tables, because they only go up to \(p=0.50\). The good news is that we can rewrite \(P(X\ge 4)\)as a probability statement in terms of \(Y\):

\(P(X\ge 4) = P(-X\le -4) = P(10 -X\le 10 - 4) = P(Y\le 6)\)

Now it's just a matter of looking up the probability in the right place on our cumulative binomial table. To find \(P(Y\le 6)\), we:

1. Find \(n=10\) in the first column on the left.
2. Find the column containing \(p=0.30\).
3. Find the 6 in the second column on the left, since we want to find \(F(6)=P(Y\le 6)\).

Now, all we need to do is read the probability value where the \(p = 0.30\) column and the (\(n = 10, y = 6\)) row intersect. What do you get?

The cumulative binomial probability table tells us that \(P(Y\le 6)=P(X\ge 4)=0.9894\). That is, the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894.

Solution

Because \(X\) is a binomial random variable, the mean of \(X\) is \(np\). Therefore, the gardener could expect, on average, \(9\times 0.80=0.72\) seeds to germinate.

What does it mean that the average is 7.2 seeds? Obviously, a seed either germinates or not. You can't have two-tenths of a seed germinating. Recall that the mean is a long-run (population) average. What the 7.2 means is... if the gardener conducted this experiment... that is, planting nine radish seeds and observing the number that germinated... over and over and over again, the average number of seeds that would germinate would be 7.2. The number observed for any particular experiment would be an integer (that is, whole seeds), but when you take the average of all of the integers from the repeated experiments, you need not obtain an integer, as is the case here. In general, the average of a discrete random variable need not be an integer.

Solution

The variance of \(X\) is:

\(np(1-p)=9\times 0.80\times 0.2=1.44\)

Therefore, the standard deviation of \(X\) is the square root of 1.44, or 1.20.